Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2011, Article ID 328394, 10 pages
doi:10.1155/2011/328394
Research Article
Existence of Positive Solutions for
Fractional Differential Equation with
Nonlocal Boundary Condition
Hongliang Gao and Xiaoling Han
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Xiaoling Han, hanxiaoling@nwnu.edu.cn
Received 29 May 2011; Revised 24 August 2011; Accepted 28 September 2011
Academic Editor: Om Agrawal
Copyright q 2011 H. Gao and X. Han. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
By using the fixed point theorem, existence of positive solutions for fractional differential equation
α
with
∞ nonlocal boundary condition D0 ut atft, ut α 0, 0 < t < 1, u0 0, u1 i1 αi uξi is considered, where 1 < α ≤ 2 is a real number, D0 is the standard Riemann-Liouville
α−1
< 1, at ∈ C0, 1, 0, ∞, ft, u ∈
differentiation, and ξi ∈ 0, 1, αi ∈ 0, ∞ with ∞
i1 αi ξi
C0, 1 × 0, ∞, 0, ∞.
1. Introduction
Fractional differential equations have been of great interest recently. It is caused both by
the intensive development of the theory of fractional calculus itself and by the applications
of such constructions in various sciences such as physics, mechanics, chemistry, and
engineering. For details, see 1–6 and references therein.
It should be noted that most of papers and books on fractional calculus are devoted to
the solvability of linear initial fractional differential equations in terms of special functions 6–
8. Recently, there are some papers that deal with the existence and multiplicity of solution or
positive solution of nonlinear initial fractional differential equation by the use of techniques
of nonlinear analysis fixed-point theorems, Leray-Shauder theory, etc.; see 9–17.
Recently, Bai and Lü 15 studied the existence of positive solutions of nonlinear
fractional differential equation
α
ut ft, ut 0,
D0
u0 u1 0,
0 < t < 1,
1.1
2
International Journal of Differential Equations
α
is the standard Riemann-Liouville differentiation, and
where 1 < α ≤ 2 is a real number, D0
f : 0, 1 × 0, ∞ → 0, ∞ is continuous.
In this paper, we study the existence of positive solutions for fractional differential
equation with nonlocal boundary condition
α
ut atft, ut 0, 0 < t < 1,
D0
∞
αi uξi ,
u0 0,
u1 1.2
i1
α
is the standard Riemann-Liouville differentiation,
where 1 < α ≤ 2 is a real number, D0
∞
and ξi ∈ 0, 1, αi ∈ 0, ∞ with i1 αi ξiα−1 < 1, at ∈ C0, 1, 0, ∞, ft, u ∈ C0, 1 ×
0, ∞, 0, ∞.
We assume the following conditions hold throughout the paper:
α−1
< 1,
H1 ξi ∈ 0, 1, αi ∈ 0, ∞ is both constants with ∞
i1 αi ξi
H2 at ∈ C0, 1, 0, ∞, at /
≡ 0 on a, b ⊂ 0, 1,
H3 ft, u ∈ C0, 1 × 0, ∞, 0, ∞.
Remark 1.1. To our knowledge, there are no results about the existence of positive solutions
for problem 1.2.
2. The Preliminary Lemmas
For the convenience of the reader, we present here the necessary definitions from fractional
calculus theory. These definitions can be found in the recent literature.
Definition 2.1. The fractional integral of order α > 0 of a function y : 0, ∞ → R is given by
α
yt I0
1
Γα
t
t − sα−1 ysds,
2.1
0
provided the right side is pointwise defined on 0, ∞.
Definition 2.2. The fractional derivative of order α > 0 of a function y : 0, ∞ → R is given
by
α
D0
yt
n t
ys
d
1
ds,
α−n1
Γn − α dt
0 t − s
2.2
where n α 1, provided the right side is pointwise defined on 0, ∞.
Definition 2.3. The map θ is said to be a nonnegative continuous concave functional on a cone
P of a real Banach space E, provided that θ : P → 0, ∞ is continuous and
θ tx 1 − ty ≥ tθx 1 − tθ y ,
for all x, y ∈ P and 0 ≤ t ≤ 1.
2.3
International Journal of Differential Equations
3
Remark 2.4. As a basic example, we quote for λ > −1,
α λ
t D0
Γλ 1 λ−α
t ,
Γλ − α 1
2.4
α α−m
t
0, m 1, 2, . . . , N, where N is the smallest integer greater than
giving in particular D0
or equal to α.
From Definition 2.2 and Remark 2.4, we then obtain the following.
Lemma 2.5. Let α > 0. If one assumes u ∈ C0, 1 ∩ L0, 1, then the fractional differential equation
α
ut 0
D0
2.5
has ut C1 tα−1 C2 tα−2 · · · CN tα−N , Ci ∈ R, i 1, 2, . . . , N, where N is the smallest integer
greater than or equal to α, as unique solutions.
Lemma 2.6. Assume that u ∈ C0, 1 ∩ L0, 1 with a fractional derivative of order α > 0 that
belongs to u ∈ C0, 1 ∩ L0, 1. Then,
α
α
D0
ut ut C1 tα−1 C2 tα−2 · · · CN tα−N ,
I0
2.6
for some Ci ∈ R, i 1, 2, . . . , N.
Lemma 2.7 see 15. Given y ∈ C0, 1 and 1 < α ≤ 2, the unique solution of
α
D0
ut yt 0,
0 < t < 1,
u0 u1 0
2.7
is
ut 1
2.8
Gt, sysds,
0
where
Gt, s ⎧
⎪
t1 − sα−1 − t − sα−1
⎪
⎪
,
⎪
⎨
Γα
0 ≤ s ≤ t ≤ 1,
⎪
⎪
t1 − sα−1
⎪
⎪
,
⎩
Γα
0 ≤ t ≤ s ≤ 1.
2.9
Lemma 2.8. Suppose H1 holds. Given y ∈ C0, 1 and 1 < α ≤ 2, the unique solution of
α
D0
ut yt 0,
u0 0,
u1 0 < t < 1,
∞
i1
αi uξi 2.10
4
International Journal of Differential Equations
is
1
ut Gt, sysds B y tα−1 ,
2.11
0
where
Gt, s ⎧
α−1
α−1
⎪
⎪
⎪ t1 − s − t − s ,
⎪
⎨
Γα
0 ≤ s ≤ t ≤ 1,
⎪
⎪
t1 − sα−1
⎪
⎪
,
0 ≤ t ≤ s ≤ 1,
⎩
Γα
∞ 1
i1 αi 0 Gξi , sysds
B y .
α−1
1− ∞
i1 αi ξi
2.12
Proof. By applying Lemmas 2.6 and 2.7, we have
ut 1
2.13
Gt, sysds C1 tα−1 C2 tα−2 .
0
Because
1
∞
αi Gξi , sds i1
∞
α−1
i1 αi ξi 1
αΓα
0
− ξi ,
αi ξiα−1 1 − ξi < αi ξiα−1 ,
2.14
1
− ξi is converge. Therefore, ∞
i1 αi 0 Gξi , sds is converge. yt is
1
continuous function on 0, 1, so ∞
i1 αi 0 Gξi , sysds is converge.
∞ 1
By u0 0, u1 ∞
i1 αi uξi , there are C2 0, C1 i1 αi 0 Gξi , sysds/1 −
∞
α−1
i1 αi ξi . Therefore,
by H1,
∞
α−1
i1 αi ξi 1
ut 1
0
B y Gt, sysds B y tα−1 ,
2.15
1
i1 αi 0
∞
Gξi , sysds
.
∞
1 − i1 αi ξiα−1
Lemma 2.9 see 15. The function Gt, s defined by 2.9 satisfies the following conditions:
1 Gt, s > 0, for t, s ∈ 0, 1,
2 there exists a positive function γ ∈ C0, 1 such that
min
Gt, s ≥ γs maxGt, s γsGs, s,
1/4≤t≤3/4
0≤t≤1
0 < s < 1.
2.16
International Journal of Differential Equations
5
Lemma 2.10 see 18. Let E be a Banach space, P ⊆ E a cone and Ω1 , Ω2 two bounded open sets of
E with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 . Suppose that A : P ∩ Ω2 \ Ω1 → P is a completely continuous operator
such that either
i Ax ≤ x, x ∈ P ∩ ∂Ω1 and Ax ≥ x, x ∈ P ∩ ∂Ω2 , or
ii Ax ≥ x, x ∈ P ∩ ∂Ω1 and Ax ≤ x, x ∈ P ∩ ∂Ω2
holds. Then, A has a fixed point in P ∩ Ω2 \ Ω1 .
Lemma 2.11 see 19. Let P be a cone in real Banach space E, Pc {x ∈ P | x ≤ c}, θ
a nonnegative continuous concave functional on P such that θx ≤ x, for all x ∈ P c , and
P θ, b, d {x ∈ P | b ≤ θx, x ≤ d}. Suppose that A : P c → P c is completely continuous, and
there exist constants 0 < a < b < d ≤ c such that
C1 {x ∈ P θ, b, d | θx > b} /
∅, and θAx > b, x ∈ P θ, b, d,
C2 Ax ≤ a, for x ≤ a,
C3 θAx > b for x ∈ P θ, b, c with Ax > d.
Then, A has at least three fixed points x1 , x2 , x3 with
x1 < a,
b < θx2 ,
a < x3 ,
2.17
θx3 < b.
Remark 2.12. If there holds d c, then condition C1 of Lemma 2.11 implies condition C3
of Lemma 2.11.
3. The Main Results
Let E C0, 1 be endowed with the ordering u ≤ v if ut ≤ vt for all t ∈ 0, 1, and the
maximum norm, u max0≤t≤1 |ut|. Define the cone P ⊂ E by P {u ∈ E | ut ≥ 0}.
Let the nonnegative continuous concave functional θ on the cone P be defined by
θu min1/4≤t≤3/4 |ut|.
Lemma 3.1 see 15. Let T : P → E be the operator defined by T ut :
then T : P → P is completely continuous.
1
0
Gt, sfs, usds,
Lemma 3.2. Let A : P → E be the operator defined by
1
Aut :
Gt, sasfs, usds B a·f·, u· tα−1 ,
0
then A : P → P is completely continuous.
3.1
6
International Journal of Differential Equations
Proof. The proof is similar to Lemma 3.1, so we omit.
Denote
⎛
M⎝
1
∞
Gs, sasds 0
N
⎞−1
Gξ
,
sasds
i
0
⎠ ,
α−1
1− ∞
i1 αi ξi
i1 αi
3/4
1
−1
γsGs, sasds
∗
.
1/4
Theorem 3.3. Assume (H1)–(H3) hold, and there exist two positive constants r2 > r1 > 0 such that
1 ft, u ≤ Mr2 , for all t ∈ 0, 1, u ∈ 0, r2 ,
2 ft, u ≥ Nr1 , for all t ∈ 0, 1, u ∈ 0, r1 , where M, N is defined in ∗,
then problem 1.2 has at least one positive solution u such that r1 ≤ u ≤ r2 .
Proof. By Lemmas 2.8 and 3.2, we know A : P → P is completely continuous, and problem
1.2 has a solution u ut if and only if u solves the operator equation u Au. In order to
apply Lemma 2.10, we separate the proof into the following two steps.
Step 1. Let Ω2 {u ∈ P | u ≤ r2 }. For u ∈ ∂Ω2 , we have 0 ≤ ut ≤ r2 for all t ∈ 0, 1. It
follows from 1 that for t ∈ 0, 1,
1
1
i1 αi 0
∞
Gξi , sasfs, usds
α−1
1− ∞
0
i1 αi ξi
⎡
⎤
∞ 1
1
i1 αi 0 Gξi , sasds
⎦
≤ Mr2 ⎣ Gs, sasds α−1
1− ∞
0
i1 αi ξi
Au ≤
Gs, sasfs, usds 3.2
r2 u.
Therefore,
Au ≤ u,
u ∈ P ∩ ∂Ω2 .
3.3
Step 2. Let Ω1 {u ∈ P | u ≤ r1 }. For u ∈ ∂Ω1 , we have 0 ≤ ut ≤ r1 for all t ∈ 0, 1. By
assumption 2, for t ∈ 1/4, 3/4, there is
Aut 1
Gt, sasfs, usds tα−1
0
≥
1
γsGs, sasfs, usds
0
≥ Nr
3/4
γsGs, sasds
1/4
r1 u.
1
i1 αi 0
∞
Gξi , sasfs, usds
α−1
1− ∞
i1 αi ξi
3.4
International Journal of Differential Equations
7
So,
Au ≥ u,
u ∈ P ∩ ∂Ω1 .
3.5
Therefore, by ii of Lemma 2.10, we complete the proof.
Example 3.4. Consider the problem
3/2
ut u2 D0
sin t 1
0,
4
5
u0 0,
0 < t < 1,
∞
u1 αi uξi ,
3.6
i1
∞
1/2
i1 αi ξi
1/5.
A simple computation showed M ≥ 1.4, N ≈ 13.6649. Choosing r1 1/70, r2 9/10, we have
where
ft, u u2 sin t 1
≤ 1.2207 ≤ Mr2 ,
4
5
ft, u u2 9
,
t, u ∈ 0, 1 × 0,
10
1
.
t, u ∈ 0, 1 × 0,
70
sin t 1 1
≥ ≥ Nr1 ,
4
5 5
3.7
With the use of Theorem 3.3, problem 3.6 has at least one positive solutions u such
that 1/70 ≤ u ≤ 9/10.
Theorem 3.5. Assume (H1)–(H3) hold, and there exist constants 0 < a < b < c such that the
following assumptions hold:
A1 ft, u < Ma for t, u ∈ 0, 1 × 0, a,
A2 ft, u ≥ Nb for t, u ∈ 1/4, 3/4 × b, c,
A3 ft, u ≤ Mc for t, u ∈ 0, 1 × 0, c, where M, N is defined in ∗.
Then, the boundary value problem 1.2 has at least three positive solutions u1 , u2 , u3 with
u1 < a,
b<
a < u3 ≤ c,
min
|u2 | < u2 ≤ c,
1/4≤t≤3/4
min
1/4≤t≤3/4
|u3 | < b.
3.8
8
International Journal of Differential Equations
Proof. We show that all the conditions of Lemma 2.9 are satisfied.
If u ∈ P c , then u ≤ c. Assumption A3 implies ft, ut ≤ Mc for 0 ≤ t ≤ 1.
Consequently,
1
1
αi 0 Gξi , sasfs, usds tα−1 ∞
i1
Au max Gt, sasfs, usds α−1
0≤t≤1 0
1− ∞
i1 αi ξi
1
tα−1
1
i1 αi 0
∞
Gξi , sasfs, usds
α−1
1− ∞
0
i1 αi ξi
⎡
⎤
∞ 1
1
α
Gξ
,
sasds
i
i
i1
0
⎦u
≤ ⎣ Gs, sasds α−1
1− ∞
α
ξ
0
i
i1
i
≤
Gs, sasfs, usds 3.9
≤ u.
Hence, A : P c → P c . In the same way, if u ∈ P a , then assumption A1 yields ft, ut <
Ma, 0 ≤ t ≤ 1. Therefore, condition C2 of Lemma 2.11 is satisfied.
To check condition C1 of Lemma 2.11, we choose ut b c/2, 0 ≤ t ≤ 1. It is
easy to see that ut b c/2 ∈ P θ, b, c, θu θb c/2 > b, and consequently,
{u ∈ P θ, b, d | θu > b} /
∅ Hence, if u ∈ P θ, b, c, then b ≤ ut ≤ c for 1/4 ≤ t ≤ 3/4.
From assumption A2, we have ft, ut ≥ Nb for 1/4 ≤ t ≤ 3/4. So,
θAu ≥
min
|Aut|
1/4≤t≤3/4
1
γsGs, sasfs, usds
0
3.10
3/4
γsGs, sasds
> Nb
1/4
b u
θAu > b, for all u ∈ P θ, b, c.
This shows that condition C1 of Lemma 2.11 is also satisfied.
By Lemma 2.11 and Remark 2.12, the boundary value problem 1.2 has at least three
positive solutions u1 , u2 , and u3 with
u1 < a,
a < u3 ,
The proof is complete.
b<
min
|u2 |,
1/4≤t≤3/4
min
|u3 | < b.
1/4≤t≤3/4
3.11
International Journal of Differential Equations
9
Example 3.6. Consider the problem
3/2
D0
ut ft, u 0,
u0 0,
u1 0 < t < 1,
∞
αi uξi ,
3.12
i1
where
∞
1/2
i1 αi ξi
1/5,
⎧ t
⎪
2
⎪
⎪
⎨ 40 14u ,
ft, u ⎪
t
⎪
⎪
u,
⎩13 40
u ≤ 1,
3.13
u > 1.
We have M ≥ 1.4, N ≈ 13.6649. Choosing a 1/14, b 1, c 36, there hold
1
,
t, u ∈ 0, 1 × 0,
14
1 3
t
u ≥ 14.025 ≥ Nb ≈ 13.7, t, u ∈
,
ft, u 13 × 1, 36,
40
4 4
ft, u ft, u ≤ 13 t
14u2 ≤ 0.097 ≤ Ma,
40
t
u ≤ 48.025 ≤ Mc ≈ 50.4,
40
3.14
t, u ∈ 0, 1 × 0, 36.
With the use of Theorem 3.5, problem 3.12 has at least three positive solutions u1 , u2 , and
u3 with
max|u1 t| <
0≤t≤1
1
,
14
1<
min
|u2 t| < max|u2 t| ≤ 36,
1/4≤t≤3/4
1
< max|u3 t| ≤ 36,
14 0≤t≤1
0≤t≤1
min
3.15
|u3 t| < 1.
1/4≤t≤3/4
Acknowledgment
This work is sponsored by the NSFC 11101335, 11026060, Gansu Provincial Department of
Education Fund 1101-02, NWNU-KJCXGC 03-69.
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